I have seen it claimed in multiple places that adding any non Clifford gate to the Clifford group yields a universal gate set. It is, however, not easy to find an accessible proof of this fact.
The question
cites Corollary 6.8.2 of Nebe et al. (Self-Dual Codes and Invariant Theory (Springer 2006)) as a proof of the fact that the Clifford group plus any other gate is universal.
This proof by Nebe makes heavy use of invariant theory and coding theory.
I'm looking for a more elementary or self contained proof of this result.
I would also be very happy with an answer along the lines of "Ian you are so silly, the proof in Nebe is actually quite elementary and self contained you just have to know blah blah blah [insert some actually clear explanation of why the ideas in the proof are elementary]."
Just to clarify, I understand the proof in Nielson and Chuang that single qubit gates together with cNot gates are universal.
And I am comfortable with the classification of the subgroups of $ PU_2=PSU_2=SO_3(\mathbb{R}), SU_2, U_2 $ in particular that the Clifford group on one qubit is a maximal closed subgroup and so adding any single qubit gate to the Clifford group gives a universal set for all single qubit gates and thus for all gates.
What I am really trying to understand is why adding any non Clifford gate, even a multi qubit gate, to the Clifford group yields a universal gate set.
Long comment: Ah you are totally right that naively taking the determinant 1 subgroup can go horribly wrong, thanks for catching that! I guess what I really meant is take the group $$ sG:=\pi_2^{-1}[\pi_1(G)]=\{ (e^{i \theta} g) \in SU(d): g \in G \} $$ where $ \pi_1: U(d) \to PU(d) $ and $ \pi_2: SU(d) \to PU(d) $ are the natural projections and $ \pi_2^{-1}[\pi_1(G)] $ just denotes taking the inverse image under $ \pi_2 $ of the group $ \pi_1(G) \subset PU(d) $ (see https://quantumcomputing.stackexchange.com/a/27232/19675 for another situation where I define and use this group $ sG $). And I'm not saying anything new because the group I describe is exactly the group you call $\overline G := \{ \det(U^\dagger)^{1/d} U \, | \, U \in G\}$ where by $ \det(U^\dagger)^{1/d} U $ you mean all possible $ d $ roots, so for example $ \det(H^\dagger)^{1/d} H $ is both $ iH $ and $ -iH $ and $ \det(I^\dagger)^{1/d} I $ is both $ I $ and $ -I $. So for $ G=\{ I, H \} \subset U(2) $ the corresponding subgroup of $ SU(2) $ is $ \{ I,-I,iH,-iH \} $ exactly as you say